Question

Integral of arcsin x

Answer 1

The last part

Answer 2

This was my approach

Answer 3

@jfernandes

Answer 4

does this fly with you?

from (iii) and FTC part 2

\( \dfrac{d}{dt}\sin^{-1} t = 2 \sqrt {1-t^2} - \dfrac{d}{dt}(t \sqrt{1-t^2}) \qquad \star\)

from (iv) and FTC part 2

\(\sqrt{1-t^2} = \dfrac{d}{dt}(t \sqrt{1-t^2}) + \dfrac{t^2}{\sqrt{1-t^2}} \qquad \triangle\)

pop \(\triangle\) into \(\star\)

\(\dfrac{d}{dt}( sin^{-1} t ) = 2 \sqrt {1-t^2} - (\sqrt{1-t^2} - \dfrac{t^2}{\sqrt{1-t^2}}) \)

\(= \sqrt {1-t^2} + \dfrac{t^2}{\sqrt{1-t^2}} \)

\(= \dfrac{1 - t^2 + t^2}{\sqrt{1-t^2}} \)

\(\implies \dfrac{d}{dt}(\sin^{-1} t ) = \dfrac{1}{\sqrt{1-t^2}} = \dfrac{d}{dt} ( \int\limits_0^t \dfrac{1}{\sqrt{1-x^2}} dx)\)

last line by FTC again

hardly ideal way to do it....

Answer 5

@irishbory123
I have a question regarding your method
\[\frac{ d }{ dx } \int\limits_{0}^{a}f(x) = f(a)\]Is this generalization correct? Can you explain how you got it?

Answer 6

@ganeshie8

Answer 7

Well, you made me look it up and its actually part **1** of the Fundamental Theorem of calculus. i called it part 2 which is wrong

It states that

\(\dfrac{d}{dx} \int\limits_a^x f(t) dt = f(x)\), where a is like a fixed point on the x axis



here it is re-stated using different letters so's the pattern is clear:

\(\dfrac{d}{dv} \int\limits_a^v f(p) dp = f(v)\)

function f has to be continuous blah blah blah

if you're interested here's patrickJMT

https://www.youtube.com/watch?v=PGmVvIglZx8

haven't watched it myself but if it's crap then he's already fooled about 850,000 other people

FTC is a good thing to know about. if you don't know it that's more evidence that there is a niftier solution to your problem....

Answer 8

Thank you very much